Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```        The general Equation of a conic section is ax^2+2hxy+by^2+2gx+2fy+c=0;

Then how did we get the discriminant as abc+2fgh-af^2-bg^2-ch^2 ??```
7 years ago

147 Points
```										Dear Moni

The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b  doesn’t variables simultaneously.

Let a ≠ 0.

Now, the above equation becomes

a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac

on completing the square on the left side, we get,

a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac.

i.e.    (ax + hy + g) = + √y2(h2–ab)+2y(gh–af) +(g2–ac)

We cannot obtain x in terms of y, involving only terms of the first  degree, unless the quantity under the radical sign be a perfect square.  The condition for this is,

(gh – af)2 = (h2 – ab) (g2 – ac)

i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc

cancelling and diving by a, we have the required  condition

abc + 2fgh – af2 – af2 –  bg2 – ch2 = 0

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it hereand we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,Askiitians ExpertsBadiuddin

```
7 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details